Answer
$$\frac{{\sqrt 3 }}{4}\pi $$
Work Step by Step
$$\eqalign{
& \int_{ - 1}^1 {\frac{{3dv}}{{4{v^2} + 4v + 4}}} \cr
& = \frac{3}{4}\int_{ - 1}^1 {\frac{{dv}}{{{v^2} + v + 1}}} \cr
& {\text{Complete the square for }}{v^2} + v + 1 \cr
& {v^2} + v + 1 = {v^2} + v + \frac{1}{4} + \frac{3}{4} \cr
& {v^2} + v + 1 = {\left( {v + \frac{1}{2}} \right)^2} + \frac{3}{4} \cr
& = \frac{3}{4}\int_{ - 1}^1 {\frac{{dv}}{{{{\left( {v + 1/2} \right)}^2} + 3/4}}} \cr
& {\text{Use the substitution method}} \cr
& u = v + 1/2,\,\,\,\,\,\,du = dv \cr
& {\text{The new limits on }}u{\text{ are found as follows}} \cr
& \,\,\,\,\,\,{\text{If }}v = 1,{\text{ then }}u = 1 + 1/2 = 3/2 \cr
& \,\,\,\,\,\,{\text{If }}v = - 1,{\text{ then }}u = - 1 + 1/2 = - 1/2 \cr
& {\text{Then}} \cr
& \frac{3}{4}\int_{ - 1}^1 {\frac{{dv}}{{{{\left( {v + 1/2} \right)}^2} + 3/4}}} = \frac{3}{4}\int_{ - 1/2}^{3/2} {\frac{{dv}}{{{u^2} + {{\left( {\sqrt 3 /2} \right)}^2}}}} \cr
& {\text{integrating by the formula }}\int {\frac{1}{{{u^2} + {a^2}}}du} = \frac{1}{a}{\tan ^{ - 1}}\left( {\frac{u}{a}} \right) + C.{\text{ with }}a = \sqrt 3 /2 \cr
& \frac{3}{4}\int_{ - 1/2}^{3/2} {\frac{{dv}}{{{u^2} + {{\left( {\sqrt 3 /2} \right)}^2}}}} = \frac{3}{4}\left( {\frac{2}{{\sqrt 3 }}{{\tan }^{ - 1}}\left( {\frac{{2u}}{{\sqrt 3 }}} \right)} \right)_{ - 1/2}^{3/2} \cr
& = \frac{{\sqrt 3 }}{2}\left( {{{\tan }^{ - 1}}\left( {\frac{{2u}}{{\sqrt 3 }}} \right)} \right)_{ - 1/2}^{3/2} \cr
& = \frac{{\sqrt 3 }}{2}\left( {{{\tan }^{ - 1}}\left( {\frac{{2\left( {3/2} \right)}}{{\sqrt 3 }}} \right) - {{\tan }^{ - 1}}\left( {\frac{{2\left( { - 1/2} \right)}}{{\sqrt 3 }}} \right)} \right) \cr
& {\text{simplify}} \cr
& = \frac{{\sqrt 3 }}{2}\left( {{{\tan }^{ - 1}}\left( {\sqrt 3 } \right) - {{\tan }^{ - 1}}\left( { - \frac{{\sqrt 3 }}{3}} \right)} \right) \cr
& = \frac{{\sqrt 3 }}{2}\left( {\frac{1}{3}\pi + \frac{1}{6}\pi } \right) \cr
& = \frac{{\sqrt 3 }}{2}\left( {\frac{\pi }{2}} \right) \cr
& = \frac{{\sqrt 3 }}{4}\pi \cr} $$
